數與座標系

設n為自然數，且n^2+6n+26為完全平方數，是求n的值為何?
Sol
n^2+6n+26
=n^2+6n+9+17
=(n+3)^2+17
(1) n^2+6n+26=(n+4)62
n^2+6n+26=n^2+8n+16
2n=10
n=5
(2) n^2+6n+26=(n+5)^2
n^2+6n+26=n^2+10n+25
4n=1(不合)
So
n=5
or
m^2=n^2+6n+26
m^2=(n+3)^2+17
m^2－(n+3)^2=17
(m－n－3)(m+n+3)=17=1*17
So
m－n－3=1，m+n+3=17
m=9，n=5

解：

n^2 + 6n + 26 為完全平方數

假設此平方數為 k^2

n^2 + 6n + 26 = k^2

n^2 + 6n + 9 + 17 = k^2

17 = k^2 - (n + 3)^2

因為17為質數只能拆成17 * 1

17 * 1 = [k + (n + 3)] * [k - (n + 3)]

k + (n + 3) = 17
k - (n + 3) = 1

兩式相減

2 (n + 3) = 16
n + 3 = 8
n = 5

希望有幫助到你

to 螞蟻

(1) n^2+6n+26=(n+4)62
這部分似乎打錯了